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15 August, 02:29

A ball is launched with an initial velocity of 28.5 m/s at a 45° angle from the top of a cliff that is 10.0 m above the water below. Use g = 9.80 m/s2 to simplify the calculations. Note: You could answer these questions using projectile motion methods, but try using an energy conservation approach instead.

(a) What is the ball's speed when it hits the water?

m/s

(b) What is the ball's speed when it reaches its maximum height?

m/s

(c) What is the maximum height (measured from the water) reached by the ball in its flight?

m

2.

A boy reaches out of a window and tosses a ball straight up with a speed of 15 m/s. The ball is 30 m above the ground as he releases it. Use energy to find the following.

(a) What is the ball's maximum height above the ground?

m

(b) What is the ball's speed as it passes the window on its way down?

m/s

(c) What is the speed of impact on the ground?

+2
Answers (1)
  1. 15 August, 02:57
    0
    When it hits the water, the initial kinetic energy will increase by additional potential energy of mgh where h is 10 m.

    Initial kinetic energy

    = 1/2 x m x (28.5) ²

    = 406.125 m

    Extra potential energy added

    = m x 9.8 x 10

    = 98m

    Total energy

    = 504.125 m J

    If v the the velocity at the bottom

    1/2 m v² = 504.125 m

    v = 31.75 m/s

    b) The velocity at the top will be equal to the horizontal component of initial velocity

    = 28.5 cos 45

    = 20.15 m / s

    c) kinetic energy at the top

    = 1/2 m x (20.15) ²

    = 203 m

    Loss of energy at the top

    = 406.125 m - 203 m

    = 203.125 m

    This energy has been converted into potential energy

    mgH = 203 m

    H = 203 / 9.8

    = 20.71 m
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